The "i" component of a vector is in the x-direction. Therefore, the x-component is 3 m/s.
Answer:
9.21954 m/s
54 m/s²
Angle is zero
Explanation:
r = Radius of arm = 1.5 m
= Angular velocity = 6 rad/s
The horizontal component of speed is given by
The vertical component of speed is given by
The resultant of the two components will give us the velocity of hammer with respect to the ground
The velocity of hammer relative to the ground is 9.21954 m/s
Acceleration in the vertical component is zero
Net acceleration is given by
Net acceleration is 54 m/s²
As the acceleration is towards the center the angle is zero.
Answer:
The acceleration of the cart is 1.0 m\s^2 in the negative direction.
Explanation:
Using the equation of motion:
Vf^2 = Vi^2 + 2*a*x
2*a*x = Vf^2 - Vi^2
a = (Vf^2 - Vi^2)/ 2*x
Where Vf is the final velocity of the cart, Vi is the initial velocity of the cart, a the acceleration of the cart and x the displacement of the cart.
Let x = Xf -Xi
Where Xf is the final position of the cart and Xi the initial position of the cart.
x = 12.5 - 0
x = 12.5
The cart comes to a stop before changing direction
Vf = 0 m/s
a = (0^2 - 5^2)/ 2*12.5
a = - 1 m/s^2
The cart is decelerating
Therefore the acceleration of the cart is 1.0 m\s^2 in the negative direction.
The only force on the system is the mass of the hoop F net = 2.8kg*9.81m/s^2 = 27.468 N The mass equal of the rolling sphere is found by: the sphere rotates around the contact point with the table.
So by applying the theorem of parallel axes, the moment of inertia of the sphere is computed by:I = 2/5*mR^2 for rotation about the center of mass + mR^2 for the distance of the axis of rotation from the center of mass of the sphere.
I = 7/5*mR^2 M = 7/5*m
Therefore, linear acceleration is computed by:F/m = 27.468 / (2.8 + 1/2*2 + 7/5*4) = 27.468/9.4 = 2.922 m/s^2
Answer:
(a) The magnitude of the lift force is 52144.71 N, approximately.
(b) The magnitude of the air resistance force opposing the movement is 17834.54 N, approximately.
Explanation:
Since the helicopter is moving horizontally at a constant velocity, we can assume that the net force acting on it is zero, then
(a) in the vertical direction we have
.
(b) Now horizontally,
